Shimrat’s theorem

I have just returned from the 8th International Symposium on Domain Theory, in Yangzhou, Jiangsu province, China. I will talk about some of the highlights of that conference below, but first, I would like to mention the following theorem, due to M. Shimrat [1]:

Theorem. Every topological space arises as a topological quotient of some T2 space, by some equivalence relation.

This was communicated to me by Marcus Tressl: he expressed some puzzlement at Lemma 8.4.5 in the book, which is directly contradicted by Shimrat’s theorem. And indeed, Lemma 8.4.5 is wrong (see the errata page), and so is the subsequent Theorem 8.4.6.

In that lemma, I claimed that if you compute the topological quotient X/≡ of a sober space X by any equivalence relation ≡, then the T0 quotient (X/≡)/=0 is sober. If that were true, and since every T2 space is sober, Shimrat’s theorem would imply that the T0 quotient of any topological space is sober, which is clearly wrong.

It took me some time to find the mistake in my argument, which naturally is obvious once you have spotted it: the sobrification functor preserves coequalizers, but that is a functor from Top to Sob, and I had reasoned as though it went from Top to Top.

The proof of Shimrat’s theorem is not hard, but clever.

The T2 space Y

We start with an arbitrary topological space X, and we form the subspace Y of X × RX consisting of those pairs (x, f) such that f(x) is rational, and f(x’) is irrational for every x’≠x.

RX is the product of X copies of the real line R. We equip R with its usual metric topology, and RX has the product topology. This way, RX is a T2 space.

What is less obvious—and this is the clever part—is that Y is T2, too, despite the fact that X, and therefore also X × RX, will in general fail to be T2.

Let us show this. Consider two distinct points (x, f) and (x’, f’) of Y. The trick is that f and f’ must be distinct. Indeed, if we had f=f’, then we cannot have x=x’, otherwise (x, f) and (x’, f’) would not be distinct. But in that case, f(x) and f’(x’) are rational, f(x’) is irrational (since x’≠x), and f(x’)=f’(x’) (since f=f’): contradiction.

Since f≠f’, we can separate them by two disjoint open neighborhoods V and V’, and therefore we can separate (x, f) and (x’, f’) by the two disjoint open neighborhoods (X × V) ∩ Y and (X × V’) ∩ Y.

The quotient map q

Now that we have established that Y is T2, we verify that the map q : Y → X, which maps every pair (x, f) in Y to x, is a quotient map. Clearly, q is surjective. The inverse image q-1(U) of every open subset U of X is (U × RX) ∩ Y, which is open, so q is continuous.

Finally, we must show that every subset A of X such that q-1(A) is open in Y is itself open in X. To this end, we show that A is a neighborhood of any of its elements.

Let x be a point in A. We fix an irrational number y, and we define f as mapping x to 0, and all other points of X to y. Note that (x, f) is in Y. Moreover, q (x, f)=x is in A, so (x, f) is in q-1(A). Since q-1(A) is open in Y, by definition of the topology of Y, we can find an open rectangle U × V in X × RX such that (x, f) ∈ (U × V) ∩ Y ⊆ q-1(A). In particular, x is in U, and it remains to show that U is included in A. For every x’ in U, we will build a function f’ in V such that f’ (x’) is rational. When this is done, (x’, f’) will be in (U × V) ∩ Y, hence in q-1(A), and that will show that x’ = q (x’, f’) is in A, as desired.

In order to construct f’, we distinguish two cases. If x’=x, we can just take f’=f. Henceforth, let us assume that x’≠x, and let us have a look at V. By definition of the product topology on RX, and since f is in V, there are finitely many points x1, …, xn in X and as many non-empty open intervals I1, …, Inin R such that f ∈ [x1 ↦ I1] ∩ … ∩ [xn↦ In] ⊆ V. (I am writing [xk↦ Ik] for the subbasic open set of all functions that maps xk to an element of Ik.) Without loss of generality, we may assume that x‘ is among x1, …, xn: if that is not the case, then add [x‘ ↦ R] to the list of subbasic open sets [xk ↦ Ik]. Reindexing if necessary, we may assume that x‘=x1. Then we define f’ as mapping x‘=x1 to any arbitrary rational point in I1, each xk with 2≤k≤n to some arbitrary irrational point in Ik, and all other elements of X to y (the irrational number we fixed at the beginning of the proof, although any other irrational number would fit). It is easy to see that f’ is in V, and that f’ (x’) is rational, and this concludes the proof. ☐

What makes the latter part of the proof work is that the rational numbers form a dense subset of R, and that its complement is also dense. One could replace R by any T2 space with a dense subset whose complement is also dense.

Highlights from ISDT 2019

As I said at the beginning of this post, I have just returned from the ISDT 2019 conference.

I am pretty happy about the two papers I had there, one with Xiaodong Jia on the algebras of the probabilistic powerdomain monad and one with Matthew de Brecht, Xiaodong Jia, and Zhenchao Lyu on so-called domain-complete and LCS-complete spaces. The first paper almost characterizes those algebras in terms of barycenters of continuous valuation, a slight variant of a notion devised by G. Choquet on measures. The second one examines a pretty large class of spaces (LCS-complete spaces, namely the Gδ subspaces of locally compact sober spaces) on which all continuous valuations extend to measures—that class includes all known classes on which that extension theorem holds, including continuous dcpos and locally compact sober spaces, but also quasi-Polish spaces and therefore in particular Polish spaces. As I have said here, I keep encountering those spaces everywhere.

Among the papers that were presented at the conference, here are some of those which made the strongest impression on me. If your paper is not in the list, that does not mean I did not enjoy it! Just like awards, any selection of papers will create a bias, making more visible some papers and less visible the remaining papers. (By the way, that bias is precisely the purpose of awards; that observation is not mine, but I cannot recall where I have seen that mentioned). Propelling some papers in the limelight is unfair to others… but that is the game.

My preferred paper at the conference is due to Kok Min Ng and Weng Kin Ho. Kok Min is a PhD student of Weng Kin’s, and has already coauthored a paper with me. Mind you, this is not the reason I think he did some great work here. He is very talented. His paper is one of the best of the conference, and he also gave the best presentation at the conference, in my opinion. What he and Weng Kin did was to extend the known characterization of continuous (Yoneda-)complete quasi-metric spaces as those quasi-metric spaces whose poset of formal balls is a continuous dcpo to the case of quasi-continuous complete quasi-metric spaces. Being lazy, I would have just defined a quasi-continuous complete quasi-metric space as one whose poset of formal balls is a quasi-continuous dcpo, but they follow a more classical (and more difficult) route: they define quasi-continuous complete quasi-metric spaces in terms of Cauchy nets, with the help of certain auxiliary predicates in the style of Kostanek and Waszkiewicz, and they characterize them as those quasi-metric spaces whose poset of formal balls is quasi-continuous and (here comes the subtle point) with a basis of finite sets {(x1, r1), …, (xn,rn)} of formal balls such that r1=…=rn. The latter is important in proofs, and at the moment I do not know whether that restriction can be lifted (I conjecture that it can).

Another paper I liked a lot is due to Hui Kou and Zhenchao Lyu, entitled “a Cartesian-closed category of domains with almost algebraic bases”. An almost algebraic basis is a basis of a continuous dcpo with an extra property of approximation from above by certain descending sequences of elements of the basis. The notion is due to Hamrin and Stoltenberg-Hansen in 2006. A primary example is obtained when X is an algebraic (Yoneda-)complete quasi-metric space: then the formal balls (x, r) where x is d-finite form such an almost algebraic basis of B(X,d). (That is lifted from Proposition 3.4 of their paper, where they prove this in the case of a complete metric space X.) All algebraic dcpos have an almost algebraic basis, and the latter property vacuously implies that the dcpo is continuous. If I remember Hui Kou’s talk correctly, the probabilistic powerdomain of a dcpo with an almost algebraic basis is again a continuous dcpo with an almost algebraic basis. What Kou and Lyu show is that the full subcategory of countably-based L-domains with an almost algebraic basis is Cartesian-closed. If we could dispense with the L-domain part, this would lead to a Cartesian-closed category of domains that would be closed under the action of the probabilistic powerdomain functor, and that would solve a longstanding open problem… (Achim Jung hates it when I call that the Jung-Tix problem, so I won’t) but of course the probabilistic powerdomain of a dcpo is almost never an L-domain. At least this shows the potential of a construction (almost algebraic bases) that apparently had been mostly forgotten.